Chapter 8, Problem 1RE

To determine

To calculate: The value of the integral ${\displaystyle {\int }_{}^{}{x}^2\sqrt{x^3-27}dx}$

Answer to Problem 1RE

Solution:

The value of the integral ${\displaystyle {\int }_{}^{}{x}^2\sqrt{x^3-27}dx}$ is $\underset{\_}{\frac{2}{9}{\left(x^3-27\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C}$

Explanation of Solution

Given:

The integral ${\displaystyle {\int }_{}^{}{x}^2\sqrt{x^3-27}dx}$

Formula used:

If the function, $u=g\left(x\right)$ has a continuous derivative on the closed interval $\left[a,b\right]$ and $f$ is continuous on the range of $g$, then

${\displaystyle {\int }_a^{b}f\left(g\left(x\right)\right)g^{\prime }\left(x\right)dx}={\displaystyle {\int }_{g\left(a\right)}^{g\left(b\right)}f\left(u\right)du}$

Calculation:

The integral can be solved by using ${\displaystyle {\int }_{}^{}{x}^{n}dx}=\frac{x^{n=1}}{n+1}$

Here $u=\left(x^3-27\right),du=3x^{2}dx$, thus it is found that

$\begin{array}{ccc}\hfill {\displaystyle {\int }_{}^{}{x}^2\sqrt{x^3-27}dx}& =& \frac{1}{3}{\displaystyle {\int }_{}^{}\sqrt{x^3-27}\left(3x^2\right)dx}\hfill \\ \hfill & =& \frac{1}{3}\frac{{\left(x^3-27\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C\hfill \\ \hfill & =& \frac{2}{9}{\left(x^3-27\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C\hfill \end{array}$